lim as x tends to infinity of (1/5)ln(x/2x+5) The answer is -ln(2/5) . I dont understand how you get this.

lim as x tends to infinity of (1/5)ln(x/2x+5) The answer is -ln(2/5) . I dont understand how you get this.

limit as x approaches 9 at (x+5)/[(sqrt of x)-3] Why does this limit not exist?

Let f(x)=x^3 when x is less than or equal to 2 and x-2 when x is greater than 2 (Piecewise function) What is the limit of x approaching 2 from the right side? Explain. I'm pretty...

So I was trying to find a solution to some unusual integral. Let say int(e^(sinx) dx). And then it came to my mind that it might exist a well defined function of x, f(x), such that lim n-> infinity...

if the limit as x approaches 5 of f(x) = 0 and lim as x approaches 5 of g(x) equals 0 the find functions of f(x) and g(x) that would make the limit as x approaches 5 of f(x)/g(x) = 10 please...

Background information: nsin(1/n) diverges using the Test for Divergence. I was playing around on my calculator and I noticed that no matter what term number you use the output...

Lt x tends to infinity sin(a/2^x)/sin(b/2^x)

Let f(x,y) = 1/(x2 - y) (a) Determine the critical points of f. (b) Does the limit of the function f at the point (0, 0) exist? Justify your answer

lim 1 1 2 3 n — [ (—)9+(—)9+(—)9+...

Show that sup(−xn) = −inf xn for any sequence (xn) in ¯ℜ (Real numbers symbol with a line on its top). Conclude that lim sup(−xn) = −lim inf xn.

A sequence xn is said to be dominated by a sequence yn if xn ≤ yn for each n. Show that, if so 1. inf xn ≤ inf yn, 2. sup xn ≤ sup yn, 3. lim inf xn...

lim x -> 0 (cos(5x) - 1)/sin(7x)

The value of limx→∞ (3^x + 3^2x)^(1/x)

limit n->infinity −(7.2^n+5.5^n/7+3.5^(n+1))?

lim x→0+ (8/x − 8/|x|)

lim x→π of 6 sin(x + sin x)

lim h->0 for ((5+h)^-1 - 5^-1)/h

limx-->2 ( sin(2x-4)/(x-2)

limx--> 0+ (1/sqrtx) - ( 1/sqrtx^2+x) Please show me steps. Im really confused.

limx-->positive infty sqrt(3x^5-2x)-sqrt(3x^5-7x) I first rationalized the whole thing. After this , i divided the top and the bottom by x^5 but the answer I got is not the...

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